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If I have a bounded linear operator $A$ on a Hilbert space $H$ whose eigenvectors form an orthonormal basis for $H$ and whose corresponding eigenvalues go to $0$ then is $A$ compact and self-adjoint?

I ask because I want to prove that $A$ defined on an orthonormal basis $\{e_k\}$ as $Ae_k=e_k/(k^2+1)$ is compact and self-adjoint. I know that it is, but I'm just wondering if appealing to the spectral theorem is valid. Thanks!

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    $\begingroup$ Yes, it's valid. In particular, we can define $A$ as the limit of finite-rank operators (with respect to the operator norm). $A$ inherits self-adjointness from the sequence since $A \mapsto A^*$ is operator-norm continuous. $\endgroup$ – Omnomnomnom May 22 '17 at 3:20
  • $\begingroup$ Thanks! Just to be clear, this holds for all $A$ with eigenvectors that form an onb and eigenvalues that go to 0 and not just for the explicit operator $A$ I gave, correct? $\endgroup$ – Robert B. Marshall May 22 '17 at 3:37
  • $\begingroup$ Yep!${}{}{}{}{}{}{}{}{}$ $\endgroup$ – Omnomnomnom May 22 '17 at 3:39
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Yes, $A $ is compact. By considering truncations of $A $ that are zero on $e_k $ for $k\geq n $, you can write $A $ as a limit of finite-rank operators.

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    $\begingroup$ Do you mean a limit of finite rank operators, or perhaps a sum of rank-one operators? $\endgroup$ – Omnomnomnom May 22 '17 at 3:21
  • $\begingroup$ Corrected, thanks. $\endgroup$ – Martin Argerami May 22 '17 at 3:41

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